Questions
A point source of electromagnetic radiation has an average power output of 800 W. The maximum value of electric field at a distance of 4.0 m from the source is
a
64.7 V/m
b
57.8 V/m
c
56.72 V/m
d
54.77V/m
detailed solution
Correct option is D
The intensity of electromagnetic wave isI=Pav2πr2=E02μ0c ∴E0=μ0cPαV2πr2 =4π×10−7×3×108×8002π×(4)2=54⋅77V/mTalk to our academic expert!
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